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Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Wed Apr 20, 2016 1:22 AM

Post by Tammy T on April 18 at 03:12:45 PM

Dear Prof. Hovasapian,

I see no information about conical, planar and angular node?
Would you please explain about these nodes? For example, what are they and what to know about them? What orbitals will have these nodes?

Thank you!

The Hydrogen Atom VII

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • The Hydrogen Atom VII 0:12
    • p Orbitals
    • Not Spherically Symmetric
    • Recall That the Spherical Harmonics are Eigenfunctions of the Hamiltonian Operator
    • Any Linear Combination of These Orbitals Also Has The Same Energy
    • Functions of Real Variables
    • Solving for Px
    • Real Spherical Harmonics
    • Number of Nodes

Transcription: The Hydrogen Atom VII

Hello, and welcome back to www.educator.com, welcome back to Physical Chemistry. 0000

Today, we are going to close out our discussion of the hydrogen atom.0004

It is going to be the 7th lesson of the hydrogen atom discussion.0008

Let us jump right on in.0011

In the last lesson, we look primarily at the S orbital, the 1S, 2S, and the S in general.0015

I will stick with black today.0022

In the last lesson, we looked primarily at the S orbitals.0026

The 1S, 2S, and the S orbitals, in general.0044

We discovered that they are spherically symmetric.0058

I will just write down here spherically symmetric.0063

What it means is that, it depends only on R, the radius of the electron from the center of the nucleus.0069

It does not depend on θ or φ.0079

Let us take at look at the P orbitals.0091

Now, we look at the P orbitals.0093

For the P orbitals, the L value is equal to 1.0104

The L is equal to 0.0107

We have a relationship between N and L and M.0111

We have the N = 2, we have L = 1, we have N = 0.0118

We have L = 1, M = + 1, and we have the L = 1, M = -1.0126

The 2P orbital has 1, 2, 3 sub orbitals.0139

You know them as PX PY and PZ.0144

The 2P orbital has three sub orbitals.0154

That is all these orbitals are, they are wave functions.0164

We call them orbitals.0171

We have something like this.0173

We have ψ 2, 1, 0 = to R 2,1 S 1,0.0174

We have ψ 2, 1, 1 that is equal to R2, 1 S1, 1.0183

We have ψ 2, 1 -1 that is R 2, 1 and S 1 -1.0190

Let us look at the S sub L superscript M.0200

Let us look at the S1, 0, S1, 1, and S 1 -1.0211

Our S1,0, which is going to be a function of θ and φ, remember these are the spherical harmonics.0215

These are the wave functions for the rigid rotator and they consist of a function of 2 variables, the θ and the φ.0223

It is going to equal 3/4 π ^½ × cos of θ and S 1, 1 of θ and φ is equal to 3/8 π ^½.0231

This is going to be sin θ E ⁺I φ.0253

Of course, we have the S1 -1, this is the function of θ and φ.0258

And this is equal 3/8 π ^½ and this time it is a sin θ but it is E ⁻I φ.0264

These are our three spherical harmonics for the N =2, L = 1.0277

We are concerning ourselves with the P orbitals, L = 1.0287

Notice, these are not spherically symmetric.0293

In other words, they depend on θ and they depend on φ.0296

This one just depends on θ but these depend on θ and φ.0299

Let us go over here.0304

I will go ahead and go to the next page.0310

These are not spherically symmetric because they depend on θ and φ.0312

We just look at the spherical harmonics but the whole wave function ψ,0343

because they depend on all three variables R, θ, and φ.0346

Notice that the S 1, 1 and the S 1-1 are complex.0355

In other words, they have that E ⁺I φ and E ⁻I φ term.0372

Let us rewrite them again, it is not a problem.0378

S 1, 1, I’m going to leave off the θ in the φ.0379

3/ 8 π ^½ sin θ E ⁺I φ and the S 1 -1 is 3/ 8 π ^½ sin of θ E ⁻I φ.0384

These are complex.0404

Recall that the S sub L superscript M, the spherical harmonics, 0407

they are Eigen functions of the Hamiltonian operator, the energy operator.0420

When we say Hamiltonian, we just mean the energy operator, the Hamiltonian operator.0436

We will also be calling it just the plain, old, energy operator.0440

The total energy, we will speak of the kinetic energy operator, the potential energy operator.0450

But when we say the energy operator, we are talking about them combine the Hamiltonian.0454

Recall that these are actually Eigen functions.0460

What that means is this.0463

You remember what Eigen function means?0465

SLM = H ̅² × L × L + 1/ 2 I SLM.0468

This right here, when we take the function, one of the spherical harmonics, 0483

when we operate on it with the Hamiltonian operator, we end up been getting the function back 0488

but this time multiply by some constant.0492

This constant is the energy Eigen value.0494

It gives us the different values of the energy that we get when we actually measure the energy.0498

This right here, this is the energy Eigen value.0504

It depends only on L, the second quantum number.0517

Notice it depends only on L, because it depends only on L, L is the same for this and this.0535

L is equal to 1, so the energies of these two are the same.0541

The energy of the S 1, 1 state = the energy of the S1 -1 state.0548

They are degenerate.0557

Since these 2 orbitals are degenerate, remember what degenerate means, they have the same energy.0561

Degenerate have the same energy, let us right that in here.0575

Since they have the same energy, we know from previous lessons that any linear combination of these orbitals also has the same energy.0583

Because the Hamiltonian operator is linear operator, we are only dealing with linear operators in quantum mechanics.0611

In any linear combination of these orbitals also has the same energy.0627

We will give the formal version.0648

More formally, we have S 1, 1 and S 1 -1 are degenerate Eigen functions of the Hamiltonian operator.0657

Therefore, some constant × S 1, 1 + some constant × S 1 -1 of any linear combination, 0683

any constant that I can put in for C1 and C2 is also and Eigen function of the Hamiltonian operator.0696

It is also an Eigen function of the Hamiltonian operator with the same Eigen value, the same energy, 0705

which is H ̅² L × L + 1/ twice the rotational inertia.0726

Let us go ahead and continue on.0740

We need to choose a linear combination, that is going to be most convenient for us.0743

We can choose because we said any linear combination will do.0752

We just need to combine in such a way that we get something that works for us.0759

Makes our lives easier, makes mathematics easier, or does something else for us that we have done.0762

We can choose any linear combination we wish.0767

Normally, the following linear combinations are the ones that we use.0782

For the P sub X orbital, we have 3P orbitals.0805

Now, we are going to actually designate them PX PY PZ.0812

The PX, we take the following linear combination ½ ^½ × S 1, 1 + S 1 -1.0816

As far as the constants are concerned, we choose 1/2¹/2, 1/√2.0831

When we add the 2, let us say the S 1 -1 multiply by this, we end up with the following.0839

We end up with 3/ 4 π ^½ sin θ cos φ.0848

Notice the what we have done here in taking the linear combination, is we have turned a complex function,0862

the 2 complex functions S 1, 1 and S 1 -1, remember they were complex.0868

They had that E and the + or –I φ term, that is complex.0873

We are doing this linear combination, we turn them into real function.0882

We got rid of the complex part.0886

It is a function of θ and φ still, but it is a real function of θ and φ.0888

That is why we took the linear combination.0892

That is the only reason to take the linear combination is because in this particular case, we wanted to deal with a real function.0896

You will see in the minute you want to deal with a real function.0902

It gives us a little bit geometrical information, as far as space is concerned.0904

That the only reason to do so.0909

It is not as though we have actually, it is not like it makes it any better than the complex functions.0911

The PY, we take the following linear combination.0919

We take - I × 1/2¹/2 and this time we take S 1, 1 – S 1 -1.0923

That is your linear combination.0934

We end up with 3/ 4 π ^½ × sin θ.0935

This time it is a sin φ.0944

What we have done here, let us move on to next one.0948

These linear combinations of the S 11 S 1 -1 linear combinations, they change.0953

S 1 1 and S 1 -1 into functions of real variables, instead of functions of real and complex variables.0969

Let us work out the P sub X in detail so you actually see where this is coming from.1010

We do not just drop this linear combination on you and say this is the wave function, deal with it.1014

Let us actually work on this out.1023

Let us workout P sub X and you could do P sub Y if you want.1026

We know exactly why it took the form that it took.1034

We know exactly where it came from.1044

We said that P sub X is equal to ½ ^½ S sub 1, 1 + S sub 1 -1.1054

That is going to be equal to 1/2 ^½, S sub 1, 1 is 3/ 8 π ^½ × the sin of θ E ⁺I φ + S sub 1 -11068

which is 3/ 8 π ^½ × the sin of θ E ^- I φ.1090

I'm going to go ahead and factor out the 3/ 8 π ^½.1103

I’m going to go ahead and factor out the sin θ, so I get the following.1108

I get ½ ^½, I get 3/8 π ^½, I get the sin θ and I'm left with E ⁺I φ + E ^- I φ.1112

I will go ahead and expand this out using Euler’s formula.1134

Let us go ahead and combine these.1144

I’m going to write this as 3/16 π¹/2 × the sin of θ.1152

This is going to be cos φ + I × the sin of φ, this is that.1166

That is going to be + cos of φ is going to be cos of φ – the sin – I × the sin of φ.1174

Remember, this is cos -φ + I × the sin – φ.1196

Take the minus sign with the angle φ.1204

Cos of - φ is cos φ, sin of - φ is - sin of φ.1206

That is where this part comes from.1214

We have I sin φ -I sin φ, that goes away.1216

We are left with 3/ 16 π ^½ × sin of θ and this is × 2 × the cos of φ.1223

When we combine the two with this, we end up with our final 3/ 4 π ^½ × sin of θ × the cos of φ.1241

There we go, we took our orbital S 1, 1 and S 1 -1.1257

We want to do something about the complex variable.1263

Because of the nature of the Eigen function, because they are both Eigen functions, 1266

any linear combination of them is also an Eigen functions.1272

We have not lost anything if we just combine them.1275

If you combine them this way, we turn a complex functions into real functions.1277

That is the only reason to take the linear combination.1283

It has to confuse why did we do this?1286

This is the only reason to help us, we do not need to deal with the complex variable.1288

We want to deal with the real variables.1292

In this case, that is the only reason we take the linear combination.1294

The linear combination you choose is absolutely your choice.1297

If for some odd reason the particular situation that you are dealing with, 1302

you need the function to look a certain way to help you do something, you can choose any linear combination you want.1305

That is the whole idea, any constant will do.1311

We have S 1, 0 is equal to 3/ 4 π ^½ and this is the cos of θ, that is the PZ.1318

S 1 1 is equal to 3/4 π ^½ sin θ cos φ, that is the PX.1341

S 1 -1 = 3/4 π ^½ sin θ sin φ, that is PY.1355

We have actually specified which one belongs to which.1369

L is equal to 1, for M sub L = 0, that is the PZ orbital.1372

For M sub L =1, that is the PX orbital.1382

For M sub L = -1, that is the P sub Y orbital.1384

Is this real spherical harmonics?1393

When we plot these, θ and φ, when we plot them in a 3 dimensional space, 1409

that one plotted give us the familiar dumbbell shaped orbitals that we remember from general chemistry.1416

You remember the shape of the P orbital.1447

One along the X axis, one along the Y axis, one a long the Z axis.1457

I’m not going to draw it.1463

That is the reason why I'm not drawing it.1464

The drawing is in your book, if you want to see them.1466

You will see in a second why is that I'm actually not drawing them.1469

Mathematically, nothing is gained by using these real functions, instead of the complex versions the E ⁺I φ, the E ⁻I φ.1473

Mathematically, nothing is gained.1488

As far as I'm concerned, nothing is gained by using these real functions for the PY and the P sub X, 1493

instead of the complex functions, which are the original derived functions.1521

The only reason they are used is because they lend the directional character or quality to the orbitals in space.1535

We want to be able to visualize an orbital.1572

By using these linear combinations in turning the spherical harmonics, at least the S 1, 1 S 1 -1 1581

which is complex into real functions, when we plot these with θ and φ, 1588

a 3 dimensional space we actually get dumbbell shaped orbitals.1594

The fact that is the only reason that this is done because it lends a directional character.1600

It allows us to wrap our mind around it, gives us a little bit of a picture of what the orbital looks like 1605

or where we might find the electron.1610

That is the only reason that they actually done this.1613

Mathematically, nothing is gained by it.1616

Because we live in space, because the real world is a 3 dimensional space, it allows us to think geometrically about these things.1618

That is the only reason that this is the case.1631

As far as chemistry is concerned, because this is a physical chemistry course, it also helps with the geometry of the molecules.1633

You remember the geometry with these real functions, 1643

the arrangement of the orbital in space correlates with a lot of what we know in terms of geometry.1648

This is always something that helps geometrical intuition.1653

There is nothing that was actually mathematically gain by doing this.1657

In fact, it is been my experience that when one deals with a complex functions as opposed 1660

to the real functions, mathematically things tend to be easier.1663

But that is just my experience.1667

My own personal commentary on this.1672

I strongly deemphasize any pictorial representation of orbitals at this level and beyond.1674

Clearly, we want to tie this into what is that you have learned in general chemistry 1710

and the shapes of the orbitals and this statistical distribution of the electron.1714

The electron cloud, things like that.1718

These pictures are still in your physical chemistry books.1721

Which is fine, it is not a problem but I strongly deemphasize that you actually think about them this way.1723

These orbitals are mathematical functions.1729

That is how you want think about it.1731

Beyond the orbitals, 1S, 2S, 2PX, 2PY, 3P, 3D, the orbitals are mathematical objects and should be considered as such.1735

I’m going to say it, I’m not going to write this out. 1773

In general, I’m a big fan of using geometry in order to help your mathematical intuition.1782

I’m a huge fan of that.1787

I think you should always use geometrical arguments, geometrical notions,1788

to help you understand what it is that is going on.1793

This is one of the few cases where I think that nothing is certainly lost by doing that but you have to be a little bit careful.1795

This is one of few exceptions where geometrical thinking, geometrical notions, 1802

I think actually get in the way of you really experiencing the power of what it is that is going on here 1808

and what it is that it is actually been done here.1818

We have been able to reduce the representation of electrons to a mathematical function.1820

When we say what is an electron, the electron is this.1824

This is the electron, it is represented by mathematical function.1827

You do not want think of it as a physical object.1831

You want to what I think it as this mathematical thing because this mathematics that 1832

you are actually going to manipulate, it is going to give you the answers that you seek.1836

For the sake of chemistry, for the sake of geometry, for the sake of the molecular orbitals 1841

that we are going to be doing later on, it is nice to have that.1846

But I would not lean on it heavily, which is why I have deemphasize pictures.1848

These are mathematical objects.1852

One final comment here, with the power of mathematics it is not constrained by physical reality.1856

It is not constrained and limited by physical reality.1876

We have to place constraints and a certain physical constraints on the things that we do, in order to do derive these,1898

these wave functions for the different quantum mechanical systems that we have been dealing with.1904

Of course, this quantum behavior as we saw, arises precisely because these particles 1908

are not free to just be wherever, move wherever.1914

We have to constrain these particles and the wave functions have to adhere to those constraints,1917

which is why quantum behavior emerges.1923

Quantum behavior exists only because of these physical constraints.1925

If the mathematics, we end up with complex functions, complex and real functions, 1929

the power of mathematics is not constrained by reality.1934

Even though we have had to use reality to derive things mathematically.1936

The moral is do not lean too heavily on pictorial representations.1942

Let the math do what the math does, and trust the math more than trust your intuition.1946

This is a direction that you want to start leaning in as you move into higher classes, move on to graduate school.1951

You want to put your thinking and become more and more abstract.1958

You want to release of the constraints on it so you can actually think in a broader sense.1961

You want to be able to think of the big picture and this is the beginning of that.1966

One final thing, I will go ahead and close off these lessons.1971

We have ψ 2, 1, 0 = R 2,1 S 1, 0.1980

We have ψ 3, 1, 0 = R 3, 1 S 1, 0.1987

We have ψ 4, 1, 0, I’m just listing some of the ones where I’m not changing the N, the primary quantum number 234, 410, is 4, 1 and is S 1, 0.1996

Notice the S 1, 0, S 1, 0, S 1, 0.2012

Recall from previous lesson, the number of nodes places where there is 0 probability 2016

for finding the electron, the number of nodes is equal to N - L - 1.2026

The 2P, it has 0 nodes.2036

2P, N is 2 L is 1, 22 -1 -1 is 0.2044

The 3P has 1 node, the 4P has 2 nodes, and so on.2051

Thank you so much for joining us here at www.educator.com.2066

We will see you next time, bye.2068